$f(n) = 3n^{2}$ $g(x) = -6x^{2}+6-2(h(x))$ $h(x) = -6x-f(x)$ $ g(h(-2)) = {?} $
Answer: First, let's solve for the value of the inner function, $h(-2)$ . Then we'll know what to plug into the outer function. $h(-2) = (-6)(-2)-f(-2)$ To solve for the value of $h$ , we need to solve for the value of $f(-2)$ $f(-2) = 3(-2)^{2}$ $f(-2) = 12$ That means $h(-2) = (-6)(-2)-12$ $h(-2) = 0$ Now we know that $h(-2) = 0$ . Let's solve for $g(h(-2))$ , which is $g(0)$ $g(0) = -6(0^{2})+6-2(h(0))$ To solve for the value of $g$ , we need to solve for the value of $h(0)$ $h(0) = (-6)(0)-f(0)$ To solve for the value of $h$ , we need to solve for the value of $f(0)$ $f(0) = 3(0^{2})$ $f(0) = 0$ That means $h(0) = (-6)(0)-0$ $h(0) = 0$ That means $g(0) = -6(0^{2})+6+(-2)(0)$ $g(0) = 6$